Can you find what is being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent for CH4 + 2O2 CO2 + 2H2O.

Question

Can you find what is being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent for CH4 + 2O2  CO2 + 2H2O.

anonymous · Answer

The equation has the correct numbers, you dont need to change anyghing.
CH4 is oxidized.
O2 is the oxidised agent
CO2 is the reducing agent I think.

anonymous · Answer

So nothing has oxidation numbers?

anonymous · Answer

CH4 + 2O2 --->  CO2 + 2H2O sorry if me forgetting the yield sign confused you

anonymous · Answer

Oh wait, I though you were talking about the equation numbers. I am not sure about the axidation numbers.

anonymous · Answer

Hm, no the oxidation numbers :/ I'm not sure. Like, would I solve the oxidation numbers for each one.. Like CH4 and then 2O2 and so on?

anonymous · Answer

I m not sure about the oxidation numbers either, sry... gtg, I have some math exams right now:-p. I think I m gonna fail so hard :-P

anonymous · Answer

its fine. and thanks a lot. Good luck! You seem really smart (: You'll do fine

anonymous · Answer

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